them. The disadvantage is in the side-thrust outwards at A and B, tending to throw down the piers or abutments. But this is a point not to be explained here, as it belongs to the subject of Mechanics. The arches in the last two propositions are said to be struck from one centre. There are, however, other kinds of arches struck from two, three, and sometimes four, centres. We will describe them in order. 202. PROP. C. To draw a two-centred arch of given span. = Let AB be the given span. With centres A and B, and radius AB, draw two arcs of circles intersecting in C. Produce AB both ways to a, and b, making Aa= Bb the depth of the archstones; and with the same centres A, and B and radius Ab, describe two other arcs bc, ac, intersecting Then the arch AC is centred to B, and BC to A, that is, the joints of the arch-stones converge to those centres respectively. The top stone, or key-stone, as it is called, of the arch, is generally, though not always, made saddleshaped, as shewn in the figure, to avoid the objectionable vertical joint at C. in c. a A B b This form of the two-centred arch is the one most commonly used, where the straight lines joining A, B, and C, form an equilateral triangle. But it is not necessary that the centres be at A and B. They may be in AB produced both ways; or the centre of BC may be any where in AB between A and the middle point; and the centre of AC between B and that middle point. The centres must be both in AB, or in that line produced, and equidistant from A and B, if not coincident with A and B. If it be desired to keep down the crown of the arch the centres will be taken between A and B. If, on the other hand, the height of the arch is to be increased, the centres will be taken in AB produced. The effect of taking the centres in AB, or AB produced, obviously is to make the arch at its lowest points A, and B, obey Rule I. (197). But it is to be observed, that Rule II. is infringed at C and c, without producing in this case any bad effect, on account of the arcs from those points each way downwards being perfectly equal and symmetrical. This arch is more easily constructed than most others, but is seldom used (except in ecclesiastical structures, for which it is supposed to possess a peculiar fitness) on account of its great height in proportion to the span. It is commonly called the pointed arch of two centres. 203. PROP. CI. To draw a three-centred arch of given span. E 1. Let AB be the given span. From each extremity cut off equal portions AC, BD, each less than half AB. With centres A and C, and radius AC, describe intersecting arcs in E: and similarly with the same radius, and centres B and D, describe intersecting arcs in F; at the same time drawing the arcs AE, BF as portions A B of the required arch. Join EC, and FD, and produce them to meet in G. Then with centre G, and radius GE, or GF, draw the arc FF; and AEFB shall be the inner boundary of the arch—that is, AEFB shall be one continuous curve. For, since the centres C and G are in the same straight line passing through the point of junction of the two arcs AE, EF, .. AE and EF have the same tangent at the point E: similarly BF, and FE have the same tangent at F. Hence AEFB is a continuous curve.-The joints of the stones from A to E are centred to C; from B to F they are centred to D; and from E to F they are centred to G. 2. Another form of a threecentred arch is constructed as follows: Bisect AB in C; with centre C and radius CA, draw equal arcs AD, BE. Join DC, EC; and produce them to meet AG, BH, which are at right angles to AB, in G and H. Then with centres G and G X H, and radius GE, or HD, draw the arcs DF, EF meeting in F. ADFEB is the inner boundary of the arch required. The stones from A to D, and from B to E, are centred to C. Those from D to F are centred to H; and those from E to F are centred to G. The proof is too obvious to need repetition. 204. PROP. CII. To draw a three-centred arch of which both the span and rise are given. In the preceding Prop. the span only was given, and we were restricted to no particular rise. Here both span and rise are fixed for us. E D F H B Let AB be the given span, and CD, at right angles to AB from its middle point C, the given rise. Upon CB as a base describe the equilateral triangle BEC, on that side of BC on which the arch is to lie. In CE take CF = CD; join A DF, and produce it to meet BE in G. Through G draw GHI parallel to EC, meeting BC in H, and DC produced in I. Make AK in AC equal to BH; join IK, and produce it indefinitely. Then with centres Hand K, and radius BH, or AK, draw arcs BG, AL; and with centre I, and radius ID, draw the arc GDL.-ALDGB shall be the inner boundary of the arch required. For, IG=ID, since CF = CD (71); and the arcs AL, DL have the same tangent at L, since their centres K, I, and the point of junction L, are in the same straight line. Also arcs BG, DG have the same tangent at G, for the same reason; .. ALDGB is one continuous curve of three centres. In constructing the arch the joints of the stones from A to L are centred to K; those from B to G are centred to H; and those from G to L are centred to I. 205. PROP. CIII. To draw a four-centred pointed arch of given span. Let AB be the given span: from each extremity mark off equal parts AC, BD, each less than half AB, with centres C and D describe arcs AE, BF, making them equal by describing intersecting arcs with the same radius from centres A and B. Join EC, FD; and produce them to meet DH, CG, which are at right angles to AB, in H and G. With centres G, H, describe the arcs FK, EK intersecting in K; and AEKFB shall be the inner boundary of the arch required. The proof is obvious from the preceding Propositions. K A I D G OBS. The above form of arch is the one most common of this class-but it is not absolutely necessary either that AEC, BFD should be equilateral triangles, or that the points G, H in FD and EC produced should be taken, and none other, for centres. Any equal arcs AE, BF, and any other points in FD and EC produced, equidistant from C and D, and beyond their point of intersection, will satisfy the Problem. 206. PROP. CIV. To draw a four-centred pointed arch of which both the span and rise are given. Let AB be the given span, and CD the given rise. Mark off equal parts, in AB, from A and B, viz. AE, BF, and with centres E and F describe equal arcs AG, BH. Join GE, HF, and produce them indefinitely. Join DG, DH; bisect GD by a straight line cutting it at right angles (101), and produce this line to meet GE produced in I. Similarly, let the line which bisects DH at right angles meet HF produced in K. K With centres I, and K describe arcs GD, HD, meeting in D. Then AGDHB is the arch required. For, that it is a continuous curve is obvious from the preceding propositions. Also, since GD is bisected by a straight line passing through the centre, .. GD is a chord of the circle (49), that is, the arc described with centre I, and radius IG, passes through D. Similarly the arc described with centre K, and radius KH, passes through D. 207. PROP. CV. To construct an Oval*, that is, a plane figure with curved symmetrical boundary returning into itself, not a circle, but composed of arcs of two or more circles forming one continuous curve. Various methods are usually given of describing an Oval, according as it is required, that the complete figure approach to, or recede from, a circle. But the following method includes them all :— Take any straight line AB, and from each extremity mark off equal parts AC, BD, each less than half AB, and greater or less, according as the oval is to approach more or less to a circle. With centres A A and C, and radius AC, describe two pairs of intersecting arcs, on opposite sides of AB, at E and F; and at the same time B draw the arc EAF. Similarly, with centres B and D, and radius BD, draw the equal arc GBH. With centres C and D, and radius CD, describe two pairs of intersecting arcs, on opposite sides of AB, at I and K. Then with centres I and K, and radius IE, draw the arcs EG, FH; and AEGBHF is the Oval required. For, joining AE, EC, IC, ID, since the triangles AEC, CDI are both equilateral, .. 4 ACE=LICD, and .. ICE is a straight line (31). Hence the arcs AE, EG have a common tangent at E. In the same manner, it will appear, that the arcs meeting at F, G and H, have also a common tangent at those points; that is, the curved line AEGBHF is continuous, as required. OBS. It is plain that the figure is divided into two equal and symmetrical parts by the line AB. Also, if a straight line be drawn through the middle point of AB, From the Latin ovum', an egg, the profile of an egg being. something like what is meant by an oval, |